Question

cot⁻¹ 1/5 +1/2 cot⁻¹12/5 =π/2,Prove it.

Answer 1

we have to prove that,

cot⁻¹ (1/5) +1/2 cot⁻¹(12/5) =π/2

Let cot^-1(1/5) = A ⇒cotA = 1/5 .....(1)

then, tanA = 1/cotA = 1/(1/5) = 5

again, let 1/2cot^-1(12/5) = B......(2)

⇒cot^-1(12/5) = 2B

⇒cot2B = 12/5

then, tan2B = 1/cot2B = 5/12

we know, tan2B = 2tanB/(1 - tan²B)

⇒5/12 = 2tanB/(1 - tan²B)

⇒5 - 5tan²B = 24tan²B

⇒5tan²B + 24tanB - 5 = 0

⇒5tan²B + 25tanB - tanB - 5 = 0

⇒(5tanB - 1)(tanB + 5) = 0

⇒tanB = 1/5 and -5

taking tanB = 1/5

hence, tanB = 1/5

tan(A + B) = (tanA + tanB)/(1 - tanA.tanB)

= (5 + 1/5)/(1 - 5 × 1/5)

= (26/5)/0 = ∞

⇒tan(A + B) = ∞

⇒A + B = tan^-1(∞) = π/2

from equations (1) and (2),

cot⁻¹ (1/5) +1/2 cot⁻¹(12/5) =π/2 [ proved]